Help needed with dropping string

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Ed de Jager
Posts: 186
Joined: Fri Jan 14, 2005 4:01 pm
Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

Thanks,i will have to buy a few 3 wattrs then.
I have changed the 10K for a 22K as a few of you said.
Here`s the new voltages:
After the 500 Ohm:405V.
After 1K:401V.
After 22K:334V.
After 2.2K:321V.
V3-1:310V.
V3-1:276.7V.
V2-1:209,2V.
V2-6:213.6V.
V1-1:208.3V.
V1-6:208V.
I had a few minutes to play the amp and it already is a completely different amp!Clean was great already but the distortion chanel is coming to life now.I already like it a lot more!
Even my kids said they liked the sound while normally they just ask me to turn the volume down.
As i said i will buy a few 3 watters tomorrow;a 150K,220K and a 470K?
Any more suggestions would be appreciated as i have to drive quite a bit to buy them:-)
One more question:the huge difference in voltage at pins 1and 6 of the PI is that normal?I measured twice but didn`t think about chaging tubes
tonelab2
Posts: 278
Joined: Wed Feb 01, 2006 11:58 pm

Re: Help needed with dropping string

Post by tonelab2 »

Ed de Jager wrote:I was talking about these resistors.Anyone familiar with them?
Sorry for the bad quality of the picture
:oops:
Ed
They look like 2watt ABs.
2watt = 18mm x 8mm
1watt= 15mm x 6mm
.5watt= 10mm x 3mm
Ed de Jager
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Joined: Fri Jan 14, 2005 4:01 pm
Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

Thanks but unfortunately they are 15x6m.
Can i use them though(since we`re not using math)?
I can`t get values above 22k in 3 watt over here.


Ed
tonelab2
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Re: Help needed with dropping string

Post by tonelab2 »

Seen plenty of 1 watters in Fenders and other amps over the years that are still going strong 30 years down the track (also the opposite ). Always a gamble with CCs. If you have the room you can use a pair in parallel or series ( if you have the right values of course) as subs till you get new ones.
Ed de Jager
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Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

What would be the worst case scenario;the 1 watter burning out or could it give more trouble then that?
Thanks in advance,Ed
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Funkalicousgroove
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Re: Help needed with dropping string

Post by Funkalicousgroove »

CC's start on fire when they burn up........that's why they're no longer made!
Owner/Solder Jockey Bludotone Amp Works
Ed de Jager
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Joined: Fri Jan 14, 2005 4:01 pm
Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

So the worst case scenario would be me being homeless.Hmm.
The good news is that i`ve found 4 220K 2 watters,gonna try em out tomorrow and let you all know.
Ed
Ed de Jager
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Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

Hi again:-)Tried out the aforementioned resistors.
The(what i thought was)220K gave 104 Volts on 1 and 6 of V1.
Two of them in series gave 141,3 Volts.More of them became to big to fit in.
You guessed it;they are 22K resistors :oops:
i wish i knew the math behind all this .Guess it`s Mouser time.

Ed
Ed de Jager
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Joined: Fri Jan 14, 2005 4:01 pm
Location: the Netherlands

update

Post by Ed de Jager »

Hi all,i tried out the following resistors:220K,330K and 470K.
They all were the AB CC`s 1 watt so i just tried them out for the sake of getting to know the effect of them.
With 220K:V1-1-190V,V1-6-182,2V.
V2-1-193,1V V2-6-199V
V3-1-313V V3-6-287V
With 330K:V1-1-190,4V V1-6-198,4V
V2-1-199,4V V2-6-205V
With 470K V1-1-199,6V V1-6-203 V
V2-1-203,6V V2-6-209,4V
V3-6-290V.
As you can see a 470K resistor is needed to bring the voltages up to specs.
I hope this is of some help to someone sometime.

Greetings,Ed
dogears
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Re: Help needed with dropping string

Post by dogears »

Hi Ed,

Remember, using a bigger resistor changes the load on the string. DUmble used the equivelent of 150K. I would try and keep it as close to 150K as possible and make a small tweak here or there to get the voltages up a tad. No way would I go bigger than your 220K. 190V is fine btw.

Lower the 22K to 20K or 18K to get the V1 and V2 voltage up a hair.
Ed de Jager
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Location: the Netherlands

Re: Help needed with dropping string

Post by Ed de Jager »

Hi Dogears,thank you VERY much for your answer.
This is my introduction to Dumble and i need all the help i can get.
The 220K is 2 watt;will that be enough?It says 3 watt in the layout i know.
The good thing also is that the resistors needed to raise those voltages i can get here locally.Isn`t the voltage on V1-6 way too low though?
For some reason shops over here aren`t carrying above 22K in 2 or 3 watt.
I know i eventually will have to buy those Dale resistors but i am going step by step :)

Ed
dogears
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Re: Help needed with dropping string

Post by dogears »

2 watt is fine. Say you have 310V there. The voltage drop is the 310V and you can easily calculate the resistor needed. I think that is .64 of a watt. 1 watt metal oxide may suffice!!
Ed de Jager wrote:Hi Dogears,thank you VERY much for your answer.
This is my introduction to Dumble and i need all the help i can get.
The 220K is 2 watt;will that be enough?It says 3 watt in the layout i know.
The good thing also is that the resistors needed to raise those voltages i can get here locally.Isn`t the voltage on V1-6 way too low though?
For some reason shops over here aren`t carrying above 22K in 2 or 3 watt.
I know i eventually will have to buy those Dale resistors but i am going step by step :)

Ed
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jelle
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Re: Help needed with dropping string

Post by jelle »

Hi Ed,

I agree with Dogears. I use 1W resistors here. Just keep 'm away from the e-caps since these do not like heat...

Good luck,

Jelle
Ed de Jager
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Re: Help needed with dropping string

Post by Ed de Jager »

Thanks guys!
The other resistors in the dropping string have to be 3 watt right?
How do you calculate these things?Is that simply Ohm`s law or is there more to it?
I have a stack of books from the 50`s that used to belong to my Father but they don`t make much sense to me.
I regret i had`t more interest in these things when he was still alive.

Ed
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heisthl
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Re: Help needed with dropping string

Post by heisthl »

Ohm's Law E(voltage) = I (current) X R (resistance). In your case you need the power formula P (power) = E X I and since you don't know the "I" restate the formula (remember your algebra) P = (E X E)/R. So for your resistor to ground just plug in the numbers, for a resistor in a string measure the voltage at each end and subtract the lower voltage from the higher and use that number for E.
Former owner of Music Mechanix
www.RedPlateAmps.com
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